>Not sure...they are supposed to email me an official
>quote. It was the single phase part, the current
>offering of the 3phase only gives the instantanous I
>and V data, not the real power numbers.
>
>Qty was 1K (smaller volumes for this project) but they
>know they are going up against ADI who's 3phase part
>is around $7. Potential problem is that I would have
>to use 3 of the chips so not sure about room on the
>board.
>
>I am however looking for a current transfomer, for up
>to 65A. Any suggestions?
>
>--- Ake Hedman <
akhespam_OUTeurosource.se> wrote:
>
>
>> >I was just quoted the other day on the Cirrus chip
>>of around $1.80 ea
>>
>>Was that for the CS5461? Quantity?
>>
>>/Ake
>>
>>alan smith wrote:
>>
>>
>>
>>>Another possibility....
>>>
>>>Use a power meter chip like from ADI or Cirrus that
>>>
>>>
>>do
>>
>>
>>>all the calculations and use a PIC to read it via
>>>
>>>
>>the
>>
>>
>>>SPI port, and the PIC then displays or whatever...
>>>
>>>I was just quoted the other day on the Cirrus chip
>>>
>>>
>>of
>>
>>
>>>around $1.80 ea
>>>
>>>---
@spam@Richard.ProsserKILLspampowerware.com wrote:
>>>
>>>
>>>
>>>
>>>>Lou,
>>>>Then for 17Amps of load (sinusiodal ?) you will
>>>>
>>>>
>>get
>>
>>
>>>>1.983Vrms accross the
>>>>56 ohm resistor.
>>>>Provided this is a sine wave, the peak voltage
>>>>
>>>>
>>will
>>
>>
>>>>be 1.414 times this or
>>>>2.8V - or 5.6V pk-pk.
>>>>
>>>>So you need to organise a 2.5V offset and some
>>>>
>>>>
>>sort
>>
>>
>>>>of voltage divider so
>>>>that the input to the ADC stays inside the 0-5V
>>>>range - with a bit of
>>>>margin and protection against turn-on transients
>>>>etc.
>>>>If the waveform is not a sinewave you may need a
>>>>significant margin. I'd
>>>>plan on an input range of no more than 3V pk-pk.
>>>>
>>>>And ensure the 56ohm load resistor is secured and
>>>>
>>>>
>>of
>>
>>
>>>>adequate rating. (3.5
>>>>x 3.5 / 56 = 220mW - Use at least a 1 watt
>>>>resistor. If it goes open
>>>>circuit the voltages will get very high.
>>>>
>>>>Probably easiest to calibrate in software but the
>>>>voltage divider could
>>>>include a pot..
>>>>
>>>>Richard P
>>>>
>>>>
>>>>
>>>>
>>>>Hello.
>>>>I'm building a power meter to measure how much AC
>>>>current some electronic
>>>>equipment is drawing (from 0 to 17 Amps).
>>>>I'm using a Triad Magnetics CSE187 current sense
>>>>transformer.
>>>>This current sense transformer rates primary
>>>>
>>>>
>>0-240V
>>
>>
>>>>0-30 AMPs and secondary
>>>>outputs 0-3.5V proportional to the current.
>>>>I have a 56 ohm resistor between the 2 transformer
>>>>secondary pins, with one
>>>>secondary pin (pin 5) to ground, and the other
>>>>secondary pin (pin 8) routed
>>>>to the analog input of a PIC18F242.
>>>>I'm using the PIC to perform the RMS current
>>>>
>>>>
>>(IRMS)
>>
>>
>>>>algorithm.
>>>>Here are my 3 questions:
>>>>Is there any circuit needed between the output of
>>>>the transformer and the
>>>>PIC analog input?
>>>>It seems to me that nothing additional is needed
>>>>because the transformer
>>>>secondary is 0-3.5V and the PIC analog is 0-5V.
>>>>Is this method accurate?
>>>>Is any calibration needed or possible?
>>>>Thank you,
>>>>Lou
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>